Q.Aqueous solutions of most $\text{Zn}^{2+}$ complexes, such as $[\text{Zn}(\text{H}_2\text{O})_6]^{2+}$, are colourless, even though $\text{Zn}^{2+}$ is a $d$-block ion. Explain why, based on its $d$-electron configuration.

[!TLDR] $\text{Zn}^{2+}$ is $d^{10}$ — every d-orbital, both sets, is already fully occupied. [!ANSWER] $[\text{Zn}(\text{H}_2\text{O})_6]^{2+}$ is colourless because $\text{Zn}^{2+}$'s $d^{10}$ configuration leave …

Zinc has the ground-state electron configuration $[\text{Ar}]3d^{10}4s^2$; losing its two 4s electrons to form $\text{Zn}^{2+}$ leaves the configuration $[\text{Ar}]3d^{10}$ — all ten d-electrons present, completely filling all five d-orbitals (the three $t_{2g}$ and two $e_g$ orbitals in an octahedral complex) with paired electrons. A $d$-$d$ transition requires an empty (or at least partially empty) higher-energy orbital for an electron to be promoted into; with the $e_g$ set already completely full, there is nowhere for an electron excited out of $t_{2g}$ to go — no such transition is physically possible. Consequently $[\text{Zn}(\text{H}_2\text{O})_6]^{2+}$ absorbs no visible light for this reason and is colourless, for essentially the opposite underlying re …

Determine the metal ion's d-electron count; a fully-filled $d^{10}$ configuration leaves no vacant higher-energy orbital for a promoted electron, exactly as a $d^0$ conf …

Assuming $d^{10}$ complexes are colourless for the same reason as $d^0$ complexes ('no d-electrons') — the correct reason is the opposite: $d^{10}$ has too MANY …